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Jet001 [13]
3 years ago
12

How many protons does a neutral nitrogen atom have

Chemistry
1 answer:
Cloud [144]3 years ago
6 0
A neutral nitrogen atom has 7 protons.
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In terms of their electron configurations, why is cesium more likely to lose its valence electron than potassium?
Harlamova29_29 [7]

Explanation:

use the term electron sheilding, the more electrons between the valence el3ctron and nucleus the easier to lose the valence electron (more sheilding = easier to lose)

5 0
3 years ago
WILL GIVE BRAINLIEST!!!
Rus_ich [418]

Answer:

The answer is B

Explanation:

The answer is B because representative particles can only be atoms.

3 0
3 years ago
From where do the placenta and umbilical cord develop?
Anastasy [175]
It’s either the first or second one

I think it’s the first one - the outer cells of the blastocyst
8 0
3 years ago
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How many carbon atoms are present in a 10.0g sample of C3H8
balu736 [363]

Answer:

4.09ₓ10²³ atoms are contained in 10g of C₃H₈

Explanation:

1 mol of C₃H₈ has 3 mol of carbon and 8 mol of hydrogen.

1 mol of C₃H₈ weighs 44 g

In 44 g of C₃H₈ we have 3 mol of C

In 10 g of C₃H₈ we'll have (10 . 3)/ 44 = 0.681 mol

1 mol of C has NA (6.02x10²³) particles

0.681 mol will have (0.681 . 6.02x10²³) = 4.09ₓ10²³ atoms

4 0
3 years ago
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77 grams of an unknown metal at 99ᵒC is placed in 225 grams of water which is initially at 22ᵒC. The water is inside a 44 gram a
BigorU [14]

Answer:

The specific heat capacity of the unknown metal is C = 0.6991 J/g°C = 0.1671 cal/g°C

Explanation:

Heat lost by the unknown metal is equal to the heat gained by the water and aluminium cup.

Given,

Mass of unknown metal = 77 g

Initial Temperature of unknown metal = 99°C

Mass of water = 225 g

Initial Temperature of water = 22°C

Mass of Aluminium cup = 44 g

Specific heat capacity of Aluminium cup = 0.22 cal/gᵒC = 0.92048 J/g°C

Final temperature of the setup = 26°C

Note that, specific heat capacity of water = 4.186 J/g°C

Let the specific heat of the unknown metal be C

Heat lost from the unknown metal

= (77)(C)(99 - 26) = (5,621C) J

Heat gained by water

= (225)(4.186)(26 - 22) = 3,767.4 J

Heat gained by Aluminium cup

= (44)(0.92048)(26 - 22) = 162.00448 J

Heat lost by unknown metal = (Heat gained by water) + (Heat gained by Aluminium cup)

5621C = 3,767.4 + 162.00448 = 3,929.40448

5621C = 3,929.40448

C = (3,929.40448 ÷ 5621) = 0.6991 J/g°C

C = 0.6991 J/g°C = 0.1671 cal/g°C

Hope this Helps!!!

6 0
3 years ago
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